I'm trying to calculate the momentum distribution of a 1D system of non-interacting identical fermions in a harmonic trap.
Given Feynman's answer (from his Statistical Mechanics book) for the position density matrix of a single trapped particle at $T>0$, $ \rho_1 (x, x'; \beta) = \sqrt{\cfrac{m \omega}{2 \pi \hbar \sinh (\beta \hbar \omega) }} \exp \left\{ \cfrac{-m \omega}{ 2 \hbar \sinh (\beta \hbar \omega) } \left[ (x^2 + x' ^2) \cosh (\beta \hbar \omega) - 2x x' \right] \right\} $ ,
the translationally invariant distribution of $ (x'-x) $ is $ \tilde{\rho} (s; \beta) = \int_{-\infty} ^\infty \mathrm{d}x \mathrm{d}x' \delta (s - (x' - x) ) \rho_1 (x, x'; \beta) = \cfrac{\mathrm{e}^{\frac{-m \omega s^2}{4\hbar} \coth \frac{\beta \hbar \omega}{2}}}{2 \sinh \frac{\beta \hbar \omega}{2}} $ .
Here, we define $\beta\equiv 1/(k_\text{B}T)$. The Fourier transform of $\tilde{\rho} (s ; \beta)$ is the momentum distribution of the system, which is also a Gaussian.
How would you construct a two-fermion version out of this? What about 3 fermions?
My first attempt was $ \tilde{\rho}_{2\text{F}} (s;\beta) = \frac{1}{2!} \left( 2 \tilde{\rho} (s; \beta) \tilde{\rho} (0; \beta) - 2 \tilde{\rho} (s; 2\beta) \right) $
using Feynman diagrams, keeping in mind the anti-periodicity $\rho_1 (x, x'; \beta + \beta) = -\rho_1 (x, x'; \beta)$ for fermions.
Visualizing the path along the imaginary time on a hypercylinder of circumference $\beta$ would look something like
Note here that diagrams with fermion paths intersecting each other have zero statistical weight.
I've written the full statement of my problem at http://mathb.in/1393 . Apparently my answer doesn't agree with two other numerical calculations... perhaps missing a normalizing factor in one of the terms.
Any comments appreciated.